WBCHSE Important Questions for Class 11 Physics For Work and Energy

Work And Energy Long Answer Type Questions

Question 1. A car is moving with a uniform velocity along a horizontal road. Does the engine of the car do any work in this instance?

The force of friction acts opposite to the direction of motion of the car. The force applied by the engine of the car against friction maintains its uniform velocity. This force is doing work as displacement occurs against friction. If f = force of friction and v = velocity of the car, work done in 1 s by the engine = force of friction x constant velocity = fv.

Question 2. Earth revolves around the sun (say in a circular path) under the action of the force exerted by the sun on the Earth. Is the sun doing any work? Explain.

Centripetal force (which is provided by the gravitational pull of the sun in this case) that acts on a body in a circular motion is perpendicular to the displacement of the body at every point of its motion and hence does not do any work. Component of this force in the direction of the displacement = Fcos90° = 0 (F = applied force). So, the gravitational pull of the sun does not do any work for the revolution of the earth, as the force acts perpendicular to the earth’s direction of displacement.

Question 3. A man is swimming against the current such that his position with respect to the river bank remains unaltered. Is the man doing any work?

The relative displacement of the man with respect to the river bank is zero. So, no work is done by the man. The resistive force of river water just balances the force applied by the man. So this case is similar to a force applied against static friction.

Question 4. When a weightlifter over his head, remains to how much stationary work does he do?

To lift a weight mg above his head through a height h, work mgh has to be done against gravity. But when the weight is held stationary above his head, no displacement occurs and no work is done by the lifter at that stage.

Question 5. A bullet is fired from a rifle. The rifle is free to recoil. Compare the kinetic energy of the rifle with that of the bullet.

K.E., \(K=\frac{1}{2} m v^2=\frac{1}{2} \frac{(m v)^2}{m}=\frac{p^2}{2 m}\) [p = momentum]

Both the rifle and the bullet were at rest before firing. After firing, the momenta of the rifle and the bullet must be equal in magnitude and opposite in direction in order to conserve linear momentum.

As p is the same for both, K ∝ \(\frac{1}{m}\). As the rifle is heavier, its K.E. is less than that of the bullet.

Question 6. If a car is driven along a straight path by an engine of constant power, find the displacement-time relation.

Let the power of the engine be P (constant).

But P = velocity x force = \(\nu m a=v m \frac{d v}{d t}\)

or, \(v d v=\frac{P}{m} d t\).

Integrating, \(\frac{v^2}{2}=\frac{P}{m} t+C\), where C= integration constant.

or, \(v=\sqrt{2} \sqrt{\frac{P}{m} t+C}\)

or, \(d s=\sqrt{2} \sqrt{\frac{P}{m} t+C} d t\left[because v=\frac{d s}{d t}\right]\)

Let \(\frac{P}{m} t+C=z\)

∴ \(\frac{P}{m} d t=d z \quad \text { or, } d t=\frac{m}{P} d z\)

∴ ds = \(\sqrt{2} \frac{m}{P} z^{\frac{1}{2}} d z\)

By further integration we get,

s = \(\frac{\sqrt{2} m \frac{2}{P}}{3} z^{3 / 2}+C^{\prime}\)

= \(\frac{\sqrt{2} m}{P} \cdot \frac{2}{3}\left(\frac{P}{m} t+C\right)^{3 / 2}+C^{\prime}\)

[C’ = integration constant]

= \(\frac{2 \sqrt{2}}{3} \sqrt{\frac{P}{m}}\left(t+\frac{m C}{P}\right)^{3 / 2}+C^{\prime}\)

This is the required relationship between time and displacement.

Question 7. A car and a lorry are moving with the same kinetic energy. Both are brought to rest with the same opposing force applied through brakes. Which one will stop within a short distance?

As soon as the lorry and the car come to rest, their kinetic energies become zero. Hence, the change in kinetic energy (= work done by the brakes) will be the same for both vehicles. Now, work done = force x displacement. As the force applied is the same, both the car and the lorry will cover the same distance before coming to rest.

Question 8. The magenta of a lighter and a heavier mass are equal. Which one of them has a greater kinetic energy?

Let the mass of the lighter body = m and its velocity = v, the mass of the heavier body = M and its velocity = V

Given, mv = MV.

Now, the kinetic energy of the lighter body, \(K_l=\frac{1}{2} m v^2\) and the kinetic energy of the heavier body, \(K_h=\frac{1}{2} M V^2\)

∴ \(\frac{K_l}{K_h}=\frac{\frac{1}{2} m v^2}{\frac{1}{2} M V^2}=\frac{m^2 v^2}{m} \times \frac{M}{M^2 V^2}=\frac{M}{m}>1\) [because m v=M V]

So, the kinetic energy of the lighter body is greater than that of the heavier body.

Question 9. The kinetic energy of a lighter body is equal to that of a heavier body. Which one of them has greater momentum?

Let the mass and velocity of the lighter body be m and v, and those of the heavier body be M and V, respectively.

Given, \(\frac{1}{2} M V^2=\frac{1}{2} m v^2\) or, \(\frac{M^2 V^2}{M}=\frac{m^2 v^2}{m}\)

or, \(\frac{m^2 v^2}{M^2 V^2}=\frac{m}{M}\)

or, \(\frac{m v}{M V}=\sqrt{\frac{m}{M}}\) or, \(\frac{\text { momentum of the lighter body }}{\text { momentum of the heavier body }}\)

= \(\sqrt{\frac{m}{M}}<1\)

Hence, the momentum of the heavier body is greater.

Question 10. A body has mechanical energy but no momentum, or has momentum but no mechanical energy—is it possible?

A body at rest at a height from the ground has potential energy, a form of mechanical energy, but no momentum.

A body with momentum must have some velocity and thus some kinetic energy associated with it. Since kinetic energy is a form of mechanical energy, a body with momentum must have some mechanical energy.

Question 11. A man with a bucket of water in his hand is standing inside a lift moving upwards. Write

  1. Whether the man does any work on the bucket of water
  2. Whether the energy of the bucket remains unchanged.


  1. As there is no displacement of the bucket with respect to the man, no work is done by the man on it.
  2. The height of the bucket of water, with respect to the ground, increases as the lift moves up. So, the potential energy of the bucket increases. Hence, the total energy of the bucket increases.

Question 12. One gets less hurt when one jumps from a height on sand than on hard floor. Why?

When a man jumps on any surface, the surface in contact with the man gets a little depressed. This depression is more in the case of a sandy surface than of a hard floor. The initial potential energy of the man is the same in both cases. So the kinetic energy just before hitting the ground is also the same.

Suppose, the mass of the man = m and he jumps from a height h. The average reaction force of the ground on him is R. He stops after depressing the ground by x. So his kinetic energy is used up in work done against the net reaction force R – mg.

As kinetic energy just before hitting the ground = potential energy at height h = mgh

(R- mg)x = mgh or, R = (1 + \(\frac{h}{x}\) mg)

As x is more for a sandy surface, R is less and hence, the man is less hurt.

Question 13. Does work done against gravity depend on the speed of lifting a body? Explain.

Magnitude of work done to raise a body through a height h,

W = weight of the body x height

Thus work done does not depend on the time of rise, i.e., on the speed of doing the work.

Question 14. Inside a compartment of a train running with a uniform velocity, a boy throws a ball. Does the kinetic energy of the ball depend on the velocity of the train?

Here, the train runs with a uniform velocity. So, if we consider the train as the frame of reference the velocity and the kinetic energy of the ball becomes zero and does not depend on the velocity of the train. When the kinetic energy is calculated, taking the earth’s surface as the inertial frame of reference, it will depend on the velocity of the train.

Question 15. To reach the same height, why is it easier to follow a path of lower inclination?

To reach the same height, the work done is equal for all paths since the potential energy attained is the same. Now, work done = force x displacement. For a path of lower inclination, this displacement is higher; so the force required is less. Thus it is easier to follow such a path.

Question 16. Show that for a simple pendulum, the work done by the tension of the string becomes zero during its oscillation.

Work done by a force is the product of the force with the component of the displacement along the force. In the case of an oscillating simple pendulum, the tension in the string always acts at right angles to the displacement of the bob. Hence, the component of displacement of the bob is zero in the direction of tension; therefore, the tension does not do any work.

Question 17. Prove that, a freely falling body of mass m loses its potential energy by \(\frac{1}{2}\)mg(2t-l) during its fall in the tth second.

Suppose the body starts from rest. The displacement in t seconds,

X = \(\frac{1}{2}\)gt

The displacement in (t – 1) seconds, y = \(\frac{1}{2}\)g(t-1)²

∴ Displacement in the t th second,

h = \(x-y=\frac{1}{2} g t^2-\frac{1}{2} g(t-1)^2=\frac{1}{2} g(2 t-1)\)

∴ Decrease in potential energy in the t th second = \(m g h=m g \cdot \frac{1}{2} g(2 t-1)=\frac{1}{2} m g^2(2 t-1)\)

Question 18. Gravitational force is a conservative force, but fric¬tional force is non-conservative —why?

A conservative force is one for which the work done can be restored. To lift a bodywork has to be done against gravity. This work gets stored in the body as its potential energy. The body, while returning to its initial state, does the same amount of work using that stored potential energy. Hence, gravitational force is a conservative force.

If the work done against a force cannot be restored, the force is called non-conservative. To pull a body over a rough surface, work has to be done against friction. This work can never be recovered. To return the body to its initial state, again work has to be done against friction. So friction is a non-conservative force.

Question 19. Can the kinetic energy of a body be negative?

A body of mass m moving with a velocity v has kinetic energy \(\frac{1}{2}\)mv². m cannot be negative, and v² being the square of a real quantity, cannot also be negative. Hence, the kinetic energy of a body cannot be negative.

Question 20. The momentum of a body is increased by 100%. What is the percentage increase in its kinetic energy?

Let the mass of the body be m.

Initial momentum = p1; hence, final momentum after a 100% increase, p2 = 2p1.

If K1 and K2 are initial and final kinetic energies, \(K_1=\frac{p_1^2}{2 m} \text { and } K_2=\frac{p_2^2}{2 m}\)

∴ \(\frac{K_2}{K_1}=\frac{p_2^2}{p_1^2}=\frac{\left(2 p_1\right)^2}{p_1^2}=4=\frac{4}{1}\)

or, \(\frac{K_2-K_1}{K_1} \times 100=\frac{4-1}{1} \times 100=300\)

So, the kinetic energy increases by 300 %.


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